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Worked solution

Derive the exponential derivative using its limiting quotient

Let f:x↦exf:x\mapsto e^x. Using the first principles of differentiation, show that f′(x)=exf^{\prime}(x)=e^x.

CJC · Lecture appendix · Example C

Factor out the fixed exponential

[f(x+h)−f(x)]/h=[ex+h−ex]/h=ex(eh−1)/h[f(x+h)-f(x)]/h=[e^{x+h}-e^x]/h=e^x(e^h-1)/h for h≠0h\ne0. Thus the required auxiliary limit is lim⁡h→0(eh−1)/h\lim_{h\to0}(e^h-1)/h.

How did this part go?

Retain the written calculator sequence

The source says Y=, enter Y1=(eX−1)/XY_1=(e^X-1)/X, then GRAPH. The graph approaches height 1 near zero but does not clearly show its missing value there. 2nd → GRAPH shows ERROR at zero. Its coarse positive-input table begins approximately 1.7183, 3.1945, 6.3618 at X=1,2,3X=1,2,3; these are screen approximations, not exact values.

How did this part go?

Set up the near-zero table

The written steps and printed setup screen specify 2nd → WINDOW, TblStart=0, ΔTbl=1E−7, with independent and dependent modes both Auto. Press 2nd → GRAPH again. The source screen displays ERROR at zero and rounded values 1 at 10−7,2×10−7,…,10−610^{-7},2\times10^{-7},\ldots,10^{-6}. The setup and rounded near-zero screen images are also used in the sine example: they are appropriate here because the same settings and displayed precision give the same screen contents. Displayed 1 does not mean an exact value of 1 at nonzero inputs.

How did this part go?

Use the stated standard limit to conclude

Page 20 states that the calculator suggests lim⁡h→0(eh−1)/h=1\lim_{h\to0}(e^h-1)/h=1. Accepting this standard limit gives f′(x)=exlim⁡h→0(eh−1)/h=exf^{\prime}(x)=e^x\lim_{h\to0}(e^h-1)/h=e^x for every real xx. Finite calculator observations motivate the limit; they are not a rigorous proof of it.

How did this part go?

Compare your sketch

Inspect the exponential quotient near zero

  1. 1. Enter the quotient

    Open Y= and enter the whole quotient using the exponential function.

    Y1
    (exp(X)-1)/X

  2. 2. Compare coarse inputs

    Open TABLE with 2nd → GRAPH. The retained source displays these coarse values.

    Compare coarse inputs
    XY1 display
    0ERROR
    11.7183
    23.1945
    36.3618
    413.4
    529.483
    667.071
    7156.52
    8372.49
    9900.23
    102202.5

  3. 3. Make the increment smaller

    Press 2nd → WINDOW for TABLE SETUP. Set TblStart=0 and ΔTbl=1E−7, with independent/dependent entries on Auto. Reopen TABLE.

    Make the increment smaller
    XY1 display
    0ERROR
    1E−71
    2E−71
    1E−61

At zero the quotient remains undefined: ERROR. The near-zero 1s are rounded displays, not exact equalities or a defined value at zero. Finite samples suggest a limit; they do not prove it. The original screens retain every intermediate table row.

View original calculator screens
Expand diagram

Use the scrollbars or arrow keys to inspect details.

Mathematical reconstruction of (exp(X)−1)/X with an open circle at (0,1), all printed coarse and near-zero calculator table values and setup start 0/step 1E−7. Zero is ERROR; near-zero values display 1 due to rounding. Original screenshots retained in source evidence.
Mathematical reconstruction of (exp(X)−1)/X with an open circle at (0,1), all printed coarse and near-zero calculator table values and setup start 0/step 1E−7. Zero is ERROR; near-zero values display 1 due to rounding. Original screenshots retained in source evidence.
Mathematical reconstruction of (exp(X)−1)/X with an open circle at (0,1), all printed coarse and near-zero calculator table values and setup start 0/step 1E−7. Zero is ERROR; near-zero values display 1 due to rounding. Original screenshots retained in source evidence.
Open original screens (new tab)

Common mistakes

A graph may hide an undefined input; rounded numerical values do not prove a limit.

Solution notes

Both originals print the complete appendix working. Source increments are written δx; this lesson uses h. No printed marks or past-paper attribution. Domain qualifications are independently authored. Appendix letters remain in the source label and identity; internal number 0 does not renumber them as printed numbered examples.

Read the hint